SQL query sorted by id's with predefined names - php

Question

I have the following SQL table
ID | country_id | name
1 33 test1
2 77 test3
3 33 test5
4 1 test6
5 77 test7
6 33 test8
I want to order this by country_id where each country_id has it's own name (but not in another database).
EX:
$countries = array(
77 => 'Macedonia',
33 => 'Uruguay',
1 => 'Ghana'
);
So the result would be something like this:
ID | country_id | name
4 1 (G hana) test6
2 77 (M acedonia) test3
5 77 (M acedonia) test7
1 33 (U ruguay) test1
3 33 (U ruguay) test5
6 33 (U ruguay) test8

Answer 1

You'll be interested in MySQL's FIELD() function which will allow you to define a custom ORDER BY query:
SELECT
id, country_id, name
FROM
countries
ORDER BY
FIELD(country_id, 77, 33, 1);
FIELD() allows you to define an "ordered list" and then checks what index/position in the list the value you pass in is located at; you can use that to generate your custom ORDER BY.
If you have an array in PHP like you showed with $countries, you could easily generate the query with:
$country_ids = join(',', array_keys($countries));
$fieldClause = 'FIELD(country_id, ' . $country_ids . ');'
Alternatively, if you have a second reference table that contains country names, you could JOIN against that:
SELECT
a.id, a.country_id, a.name
FROM
your_table a
JOIN countries c
on a.country_id = c.id
ORDER BY
c.name;
I would recommend the latter approach since it's more portable.

Answer 2

You need to use the $countries array, sort it and use it in the ORDER BY:
sort($countries);
$order = implode(', ', array_keys($countries));
Now the array is sorted by the values and you can use the keys as the order in FIELD() as shown by newfurniturey:
SELECT id, country_id, name
FROM countries
ORDER BY FIELD(country_id, $order);
However, even with only 3 countries I would create another table and join that.